Elasticity
Question:
A telephone wire 120m long and 2.2mm in diameter is stretched by a force of 380N. What is the longitudinal stress? If the length after stretching is 120.10m, what is the longitudinal strain? Determine Young’s modulus for the wire.
Solution:
The cross-sectional area of the wire is……
A = πD² / 4
= π (2.2 x 10^-3 m)² / 4
= 3.8 x 10^-6 m²
Stress = F/A
= 380 N / (3.8 x 10^-6 m²)
= 100 x 10^6 N/m²
= 100 MPa
Strain = ∆l / l
= 0.01 m / 120 m
= 8.3 x 10^-4
Y = stress / strain
= 100 MPa / (8.3 x 10^-4)
= 120,000 Mpa
Question
The density of aluminum is 2.70g/cm³. What volume does 2.0kg occupy?
Solution:
Ρ = m/V or V = m/ρ
= 2000g / 2.70g/cm³
= 741 cm³
Question
The platform is suspended by four wires at its corners. The wires are 3m long and have a diameter of 2.0mm. Young’s modulus for the material of the wires is 180000Mpa. How far will the platform drop (due to elongation of the wires) if a 50kg load is placed at the center of the platform?
Solution:
• ∆L = (3m)(123N) / (3.14x10^-6m²)
= 65x10^-5m
= 0.65mm
Question
Determine the fractional change in volume as the pressure of the pressure of the atmosphere (0.1Mpa) around a metal block is reduced to zero by placing the block in vacuum. The bulk modulus for the metal is 125000 Mpa.
Solution:
B = ∆V / V
= -∆p / B
= - (-0.1) / 125000
= 8x10^-7
Question
Compute the volume change of solid copper cube, 40mm on each edge, when subjected to a pressure of 20 Mpa. The bulk modulus for copper is 125000Mpa.
Solution:
∆V = -V∆p / B
= -(40mm)³(20Mpa) / 125000Mpa
= -10mm³
Question
To inspect a 14,500 Ncar,it is raised with a hydraulic lift.if the radiusof the small piston is 4.0cm,and the radius of the large piston is 17cm, find the force that must be exerted on the small piston to lift the car.
Solution
F₁ = F₂(A₁/A₂)
= (14,500 N)(π(0.040m)²/(π(0.17m)²)
= 800 N
Sunday, January 17, 2010
ELASTICITY
Question:
A telephone wire 120m long and 2.2mm in diameter is stretched by a force of 380N. What is the longitudinal stress? If the length after stretching is 120.10m, what is the longitudinal strain? Determine Young’s modulus for the wire.
Solution:
The cross-sectional area of the wire is……
A = πD² / 4
= π (2.2 x 10^-3 m)² / 4
= 3.8 x 10^-6 m²
Stress = F/A
= 380 N / (3.8 x 10^-6 m²)
= 100 x 10^6 N/m²
= 100 MPa
Strain = ∆l / l
= 0.01 m / 120 m
= 8.3 x 10^-4
Y = stress / strain
= 100 MPa / (8.3 x 10^-4)
= 120,000 Mpa
Question
The density of aluminum is 2.70g/cm³. What volume does 2.0kg occupy?
Solution:
Ρ = m/V or V = m/ρ
= 2000g / 2.70g/cm³
= 741 cm³
Question
The platform is suspended by four wires at its corners. The wires are 3m long and have a diameter of 2.0mm. Young’s modulus for the material of the wires is 180000Mpa. How far will the platform drop (due to elongation of the wires) if a 50kg load is placed at the center of the platform?
Solution:
· ∆L = (3m)(123N) / (3.14x10^-6m²)
= 65x10^-5m
= 0.65mm
Question
Determine the fractional change in volume as the pressure of the pressure of the atmosphere (0.1Mpa) around a metal block is reduced to zero by placing the block in vacuum. The bulk modulus for the metal is 125000 Mpa.
Solution:
B = ∆V / V
= -∆p / B
= - (-0.1) / 125000
= 8x10^-7
Question
Compute the volume change of solid copper cube, 40mm on each edge, when subjected to a pressure of 20 Mpa. The bulk modulus for copper is 125000Mpa.
Solution:
∆V = -V∆p / B
= -(40mm)³(20Mpa) / 125000Mpa
= -10mm³
Tuesday, January 12, 2010
Hydrostatic.
tutorial
1. Find the pressure due to the fluid at a depth of 76 cm in still
Water (ρw = 1.00g/cm^3)
Mercury (ρ = 13.6g/cm^3)
Solution :
P = ρgh
= (1000 kg/m^3) (9.8m/s^2) (0.76)
= 7.448kPa @ 7.5kPa
P = ρgh
= (13600 kg/m^3) (9.8m/s^2) (0.86)
= 1.01 x 10^5Nm^2 @ 101.3 x 10^3N/m^2
2. A weight piston confines a find density ρ in a closed continue. The combined weight of piston and weight iss 200N, the total cross-sectional area of the piston is A = 8.0cm^2. Find the total pressure at point B if the fluid is mercury and h = 25cm. What would and ordinary pressure gauge read at B?
Solution :
P = (F/A) + ρgh
= (2000/8 x 10^ -4) + (13 600) (9.81m/s^2) (25 x 10^ -2)
= 250 000 + 33 354
= 2.8 x 10^5 Pa
P m = (13.6g/cm^3) (9.81m/s^2) (76cm)
= (0.136kg/m) (9.81m/s^2) (0.76cm)
= 1.01 x 10^5N/m
P = (2.8 x 10^5Pa) + Pm
= 92.8 x 10^5Pa) + (1.01 x 10^3N/m)
= 3.8 x 10^5Pa
3. A vertical test tube has 2.0cm of oil (ρ = 0.8g/cm^2) floating on 8.0cm of water. What is the pressure at the bottom of the tube due to the fluid in it?
Solution :
P1 = ρgh
= (800kg/m) (9.81m/s^2) (2 x10^ -2)
= 156.96
P2 = (1000) (9.81m/s^2) (8 x 10^ -2)
= 784.8
P1+P2 = 156.96 + 784.8
= 0.94kPa
4. The U-tube device connected to the tank. What is the pressure in the tank if atmosphera ρ is 76cm of mercury? The density of mercury is 13.6kg/m^3?
Solution :
P = (76cm) – (13.6g/cm^3) (9.81m/s^2) (5 x 10^ -2)
= (76cm) – (13 600kg/cm^3) (9.81m/s^2) (5 x 10^ -2)
= 94.65 x 10^3
= 95kPa
5. The mass of a block of a aluminium is 25.0g.
What is its volume?
What will be tthe tension in a string that suspends the block when the block is totally submerged in water? The density of aluminium is 2700kg/m^3.
Solution :
V = m/p
= (0.025) / (2700)
= 9.26cm^3
ft = v,( l-Pal )
= (9.26 x 10^ -6) (9.81) (1000-2700)
= 0.154N
6. A solid aluminium cylinder with ρ=2700kg/m^3 has a measured mass of 67g in air and 45g when immersed in turpentine. Determine the density of turpentine?
Solution :
Pt = RB / VTg
FB+FR = mg
FB = mg – FR
= (0.067 x 9.81) – (0.045 x 9.81)
= 0.21582
Vr = m / p
= (0.067) / (2700)
= 2.48 x 10^ -5
Pt = 0.21582
= (2481 x 10^ -5) (9.81)
= 886.7
= 8.9 x 10^2 kg/m^3
1. Find the pressure due to the fluid at a depth of 76 cm in still
Water (ρw = 1.00g/cm^3)
Mercury (ρ = 13.6g/cm^3)
Solution :
P = ρgh
= (1000 kg/m^3) (9.8m/s^2) (0.76)
= 7.448kPa @ 7.5kPa
P = ρgh
= (13600 kg/m^3) (9.8m/s^2) (0.86)
= 1.01 x 10^5Nm^2 @ 101.3 x 10^3N/m^2
2. A weight piston confines a find density ρ in a closed continue. The combined weight of piston and weight iss 200N, the total cross-sectional area of the piston is A = 8.0cm^2. Find the total pressure at point B if the fluid is mercury and h = 25cm. What would and ordinary pressure gauge read at B?
Solution :
P = (F/A) + ρgh
= (2000/8 x 10^ -4) + (13 600) (9.81m/s^2) (25 x 10^ -2)
= 250 000 + 33 354
= 2.8 x 10^5 Pa
P m = (13.6g/cm^3) (9.81m/s^2) (76cm)
= (0.136kg/m) (9.81m/s^2) (0.76cm)
= 1.01 x 10^5N/m
P = (2.8 x 10^5Pa) + Pm
= 92.8 x 10^5Pa) + (1.01 x 10^3N/m)
= 3.8 x 10^5Pa
3. A vertical test tube has 2.0cm of oil (ρ = 0.8g/cm^2) floating on 8.0cm of water. What is the pressure at the bottom of the tube due to the fluid in it?
Solution :
P1 = ρgh
= (800kg/m) (9.81m/s^2) (2 x10^ -2)
= 156.96
P2 = (1000) (9.81m/s^2) (8 x 10^ -2)
= 784.8
P1+P2 = 156.96 + 784.8
= 0.94kPa
4. The U-tube device connected to the tank. What is the pressure in the tank if atmosphera ρ is 76cm of mercury? The density of mercury is 13.6kg/m^3?
Solution :
P = (76cm) – (13.6g/cm^3) (9.81m/s^2) (5 x 10^ -2)
= (76cm) – (13 600kg/cm^3) (9.81m/s^2) (5 x 10^ -2)
= 94.65 x 10^3
= 95kPa
5. The mass of a block of a aluminium is 25.0g.
What is its volume?
What will be tthe tension in a string that suspends the block when the block is totally submerged in water? The density of aluminium is 2700kg/m^3.
Solution :
V = m/p
= (0.025) / (2700)
= 9.26cm^3
ft = v,( l-Pal )
= (9.26 x 10^ -6) (9.81) (1000-2700)
= 0.154N
6. A solid aluminium cylinder with ρ=2700kg/m^3 has a measured mass of 67g in air and 45g when immersed in turpentine. Determine the density of turpentine?
Solution :
Pt = RB / VTg
FB+FR = mg
FB = mg – FR
= (0.067 x 9.81) – (0.045 x 9.81)
= 0.21582
Vr = m / p
= (0.067) / (2700)
= 2.48 x 10^ -5
Pt = 0.21582
= (2481 x 10^ -5) (9.81)
= 886.7
= 8.9 x 10^2 kg/m^3
Elasticity.
tutorial.
1. An iron rod 4.00m long an 5.00cm^2 in cross section stretches 1.00mm. When a mass of 22.5kg is hung from its lower end. Compute Young’s Modulus from the iron.
Solution :
σ= F/A
= (225kg) (9.8m/s^2) / (0.500 x 10^ -4)
= 4.41 x 10^7 Pa
ε = Δƪ / ƪ˳
= (1.00 x 10^ -3 m) / (4.00m)
= 2.5 x 10^ -4
Y = σ / ε
= (4.41 x 10^7 Pa) / (2.5 x 10^ -4)
= 1.764 x 10^11
= 176 GPa
2. A load of 50 kg is applied to the lower end of a steel rod 8cm long and 0.60cm in diameter. How much will the rod sterch Y = 190 GPa for stell?
Solution :
Formula : Y = F ƪ˳ / AΔ ƪ A ƪ = F ƪ˳ / Y. A
m = 50 kg x 9.81
d = 0.6 cm / 2
= 0.3
j^2 = 3 x 10^ -3m
A = п j^2
= п (3 x 10^ -3)^2
= 2.83 x 10^ -5
ƪ˳ = 80 cm
= 80 x 10^ -2
= 0.8
Y = 190 GPa
Aƪ˳ = F ƪ˳ / YA
= (490.5) (0.8) / (190 x10^9) (2.83 x 10^5)
= 392.4 / 5.377 x 10^8
= 72.97 x 10^ -7
= 73 x 10^ -6
= 73μm
3. A platform is suspended by four wire at its corner. The wire are 3.0m long and have a diameter at 2.0mm Young’s Modulus for the material is 180GPa. How for will the platform drop (due to elongation of the wire) if a 50kg load is placed at the center of the platform?
Solution :
ƪ˳ = 3m
F = 50 x 9.81
= 490.5N
A = пj^2
= (3.14 x 10^ -6)m
A ƪ˳ = 490.5Nm / (5.652 x 10^ -4)
= 2.6 x 10^6
= 0.65m
4. Two parallel and opposite force. Each 400N are applied tangentially to the upper and lower forces of a cubical matel block 25cm on a side. Fine the angle of shear and the displacement of the upper surface relative to the surface. The shear modulus for the metal is 80GPa.
Solution :
F = 400N
h = 25cm
A = 0.0625m^2
S = 80 x 10^4 p
tanθ = (2 x 10^ -7) / (25 x 106 -2)
= 8.0 x 10^ -7 rad
x = FAh / SA
x = (4000) (0.25 x 10^ -3) / (8.0 x 10^6Pa) (0.0625)
= 2 x 10^ -7m
5. The bulk modulus of water is 2.1GPa. Compute the volume contraction of 100ml of water when subjected to a pressure of 1.5mPa?
Solution :
Av˳ = - Pv˳ / B
B = - Pv˳ / Δ v˳
B = 2.1GPa
P = 1.5MPa
= 1.5 x 10^6Pa
v˳ = (100ml) / (1000)
Δ v˳ = - Pv˳ / B
= (1.5 x 10^6) (0.1) / (2.1 x 10^9)Pa
= (-7.14 x 10^ -5mm) x (1000)
= -0.0714ml
6. The compressibility of water is 5.0 x 10^ -10m^2/N. Find the decrease in volume of water of 100ml when subjected to a pressure of15mPa?
Solution :
A v˳ = B x v˳
= (15 x 10^6) (5.0 x 10^ -10Nm^ -2)
= 7.5 x 10^ -3) x(100)
= 0.75ml
1. An iron rod 4.00m long an 5.00cm^2 in cross section stretches 1.00mm. When a mass of 22.5kg is hung from its lower end. Compute Young’s Modulus from the iron.
Solution :
σ= F/A
= (225kg) (9.8m/s^2) / (0.500 x 10^ -4)
= 4.41 x 10^7 Pa
ε = Δƪ / ƪ˳
= (1.00 x 10^ -3 m) / (4.00m)
= 2.5 x 10^ -4
Y = σ / ε
= (4.41 x 10^7 Pa) / (2.5 x 10^ -4)
= 1.764 x 10^11
= 176 GPa
2. A load of 50 kg is applied to the lower end of a steel rod 8cm long and 0.60cm in diameter. How much will the rod sterch Y = 190 GPa for stell?
Solution :
Formula : Y = F ƪ˳ / AΔ ƪ A ƪ = F ƪ˳ / Y. A
m = 50 kg x 9.81
d = 0.6 cm / 2
= 0.3
j^2 = 3 x 10^ -3m
A = п j^2
= п (3 x 10^ -3)^2
= 2.83 x 10^ -5
ƪ˳ = 80 cm
= 80 x 10^ -2
= 0.8
Y = 190 GPa
Aƪ˳ = F ƪ˳ / YA
= (490.5) (0.8) / (190 x10^9) (2.83 x 10^5)
= 392.4 / 5.377 x 10^8
= 72.97 x 10^ -7
= 73 x 10^ -6
= 73μm
3. A platform is suspended by four wire at its corner. The wire are 3.0m long and have a diameter at 2.0mm Young’s Modulus for the material is 180GPa. How for will the platform drop (due to elongation of the wire) if a 50kg load is placed at the center of the platform?
Solution :
ƪ˳ = 3m
F = 50 x 9.81
= 490.5N
A = пj^2
= (3.14 x 10^ -6)m
A ƪ˳ = 490.5Nm / (5.652 x 10^ -4)
= 2.6 x 10^6
= 0.65m
4. Two parallel and opposite force. Each 400N are applied tangentially to the upper and lower forces of a cubical matel block 25cm on a side. Fine the angle of shear and the displacement of the upper surface relative to the surface. The shear modulus for the metal is 80GPa.
Solution :
F = 400N
h = 25cm
A = 0.0625m^2
S = 80 x 10^4 p
tanθ = (2 x 10^ -7) / (25 x 106 -2)
= 8.0 x 10^ -7 rad
x = FAh / SA
x = (4000) (0.25 x 10^ -3) / (8.0 x 10^6Pa) (0.0625)
= 2 x 10^ -7m
5. The bulk modulus of water is 2.1GPa. Compute the volume contraction of 100ml of water when subjected to a pressure of 1.5mPa?
Solution :
Av˳ = - Pv˳ / B
B = - Pv˳ / Δ v˳
B = 2.1GPa
P = 1.5MPa
= 1.5 x 10^6Pa
v˳ = (100ml) / (1000)
Δ v˳ = - Pv˳ / B
= (1.5 x 10^6) (0.1) / (2.1 x 10^9)Pa
= (-7.14 x 10^ -5mm) x (1000)
= -0.0714ml
6. The compressibility of water is 5.0 x 10^ -10m^2/N. Find the decrease in volume of water of 100ml when subjected to a pressure of15mPa?
Solution :
A v˳ = B x v˳
= (15 x 10^6) (5.0 x 10^ -10Nm^ -2)
= 7.5 x 10^ -3) x(100)
= 0.75ml
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